1. to add the 1st segment B+C, obvious and enough to draw an arc center on B with radius 1, which intersects at the point +C the vertical to the half of AB, as well as the arc from A radius phi at the point D.
2. from D clockwise to +E, and back to counterclockwise to F, thanks to an arc from A, symmetrical to the previous on B, radius = 1, intersecting in F the arc from B with radius BA.

3. from F clockwise to +G, then counterclockwise to H,
4. it's easy now,
again to close a triangle (+GH+I) and open another one (H+IJ)

 5. Need a fifth figure?
